PHP - Query String Into Single Value?

I'm so sorry for this question but I not really know how to play with single and double quote.

If I have a query like this:

Code: [Select]
  mysql_query('UPDATE table SET Status=1,Sending=Done WHERE ID IN ('.implode(',', $done).')');
And I wish to add Code: [Select]
SentAt='$date' in the query as well , and I try this:

Code: [Select]
  mysql_query('UPDATE table SET Status=1,Sending=Done,SentAt='$date' WHERE ID IN ('.implode(',', $done).')');
Not working...how should I write it?

Thank you.

I have a table that contains the schools in my system:

schools
id
name
location
...

Then, I have three tables that use the id of this table:

schoolAdmins
schoolID

schoolContests
schoolID

students
schoolID

When I go to delete a school from my system, I want to check to see if that school is connected to any of these three other tables first.  This is what I tried (but obviously failed because I'm here) where I'm passing the query the $studentID in question:

SELECT * FROM schoolAdmins, schoolContests, students WHERE (schoolAdmins.schoolID = $schoolID) OR (schoolContests.schoolID = $schoolID) OR (students.schoolID = $schoolID)

I'm really new to the concept of querying multiple tables in a single statement, so I'm just kind of guessing at this point.  Thanks in advance.

Hey, I was wondering if there is a way to pull multiple rows at once using a list of unique identifiers. For example, I want to pull the rows with the IDs of 4,13,91 and 252

I know the WHERE part of this query is incorrect, but I'm putting it to hopefully help you guys understand what I'm looking for.

$result = mysql_query("SELECT * FROM $table WHERE id='4' OR '13' OR '91' OR '252'");
while($row = mysql_fetch_array($result))
{
echo($row['name']);
}


Or is the best way simply to do it one query at a time without a while statement? There could be as many as a few dozen records being pulled per page.

I'm going to create a web page with <iframe> tag & want that the src of <iframe> change as 

I've made a few attempts at this, but I'm likely not making much sense.... I'm working on how to phrase the question.

I had a MySQL database where I used to combine user first AND last names in a single string... I called that string "Member"

So if I filtered:
www.mysite.com/members.php?Member=Peter

It would find and display all Peters: Peter Rabbit and Peter Griffith and Pope Peter.  This is what I want.

I'm now keeping each name (first and last) in separate strings (as separate variables?), and I am combining them in an array to display on the website:

Code: [Select]
$First = $First;
$Last = $Last;

$member = array($Last, $First,);
foreach ($Member as $key => $v ) if (!$v) unset ($Member[$key]);
$Member = implode(', ', $Member);
echo $Member;

Now I've lost the ability to search all the members for Peter.

So if I filtered:
www.mysite.com/members.php?First=Peter
...it would work.... for Peter Rabbit and Peter Griffith, but not Pope Peter

I cannot query the entire array?  I can no longer check members (first and last names) for Wayne?
www.mysite.com/members.php?Member=Peter

I'm betting I can, but there's a lot more to it... and 8 hours of Googling haven't helped much.

Thanks for listening.

~Wayne

I have seen forums highlight what we searched on google to get to their site.

is there a script that would allow me do the same???

Hi there,
im trying to have a form show up when user clicks "add joke". I need the variable to be retrieved from the url query string. I cant get the form to show up.  I think its either an issue with the GET function at the top or the link down at the bottom. Please help!


<?php
  // If the user wants to add a joke
  
    $_GET['addjoke'] = $addjoke;

  if (isset($addjoke)):
?>

<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST>
<P>Type your joke he <BR>
<TEXTAREA NAME="joketext" ROWS=10 COLS=40 WRAP>
</TEXTAREA><BR>
<INPUT TYPE=SUBMIT NAME="submitjoke" VALUE="SUBMIT">
</FORM>

<?php
  else:

    // Connect to the database server
    $dbcnx = @mysql_connect("servername", "username", "password");
    if (!$dbcnx) {
      echo( "<P>Unable to connect to the " .
            "database server at this time.</P>" );
      exit();
    }

    // Select the jokes database
    if (! @mysql_select_db("jhodara2") ) {
      echo( "<P>Unable to locate the joke " .
            "database at this time.</P>" );
      exit();
    }

    // If a joke has been submitted,
    // add it to the database.

$joketext = $_POST['joketext']; 
$submitjoke = $_POST['submitjoke']; 

    if ("SUBMIT" == $submitjoke) {
      $sql = "INSERT INTO jokes SET " .
             "JokeText='$joketext', " .
             "JokeDate=CURDATE()";
      if (mysql_query($sql)) {
        echo("<P>Your joke has been added.</P>");
      } else {
        echo("<P>Error adding submitted joke: " .
             mysql_error() . "</P>");
      }
    }
  
    echo("<P> Here are all the jokes " .
         "in our database: </P>");
  
    // Request the text of all the jokes
    $result = mysql_query(
              "SELECT JokeText FROM jokes");
    if (!$result) {
      echo("<P>Error performing query: " .
           mysql_error() . "</P>");
      exit();
    }
  
    // Display the text of each joke in a paragraph
    while ( $row = mysql_fetch_array($result) ) {
      echo("<P>" . $row["JokeText"] . "</P>");
    }
  
    // When clicked, this link will load this page
    // with the joke submission form displayed.
    echo("<P><A HREF='$PHP_SELF?addjoke=1'>Add a Joke!</A></P>");
  
  endif;
  
?>


see the problem live at http://www.freewaycreative.com/insert2.php

Hello everyone.

I have a small problem.  I might receive a ?aff=## or not on the end of my url when I get a visitor to my website.  This depends on if they are sent from a affiliate website or not.  If they are it shows fine on the home page but I loss it if they go to another page on my website.  I need to keep this ?aff=## information while they look at the other pages.  How would I capture and pass this information to my other pages as they surf my site? 
I tried this to no avail.
$aff=$_GET['aff'];
I have no clue no adding it back to the next page they go to.  Any ideas would be helpful..

Is it possible to add a query string for example some_var=jk84 to any sort of link be it, http://www.website.com , http://www.website.com/?some_id=4, or http://www.website.com/post=45&category=9 or http://www.website.com/somepost/ ?

While adding that extra query string how can I make sure I'm not affecting the website's content or causing some script error?

Hi guys,   I am using this code to open and close a pop up window, but as soon as i click the close button this   http://localhost/popup.php?random=&button=   automatically adds in the url, Please tell me what is wrong with the script

<script type="text/javascript">

 $(document).ready(function(){
    $('a.popup-window').click(function(){
       
       var popupBox = $(this).attr('href');
       $(popupBox).fadeIn(400);
	   
	   var popMargTop = ($(popupBox).height() + 24)/2;
	   var popMargLeft = ($(popupBox).width() + 24)/2;
	   
	   $(popupBox).css({
		   'margin-top' : -popMargTop,
		   'margin-left' : -popMargLeft
	   });
	   
	   $('body').append('<div id="mask"></div>');
	   $('#mask').fadeIn(400);   
	   return false;
	});
    
	$('button.close,#mask').live('click', function(){
		   $('#mask,.popupInfo').fadeOut(400,function(){
			   $('#mask').remove();
			   });	
			   return false;
			});	  	
    
       
 
 });
 $(document).keyup(function(e){
	 if(e.keyCode ==27){
		 $('#mask,.popupInfo, #popup-box').fadeOut(400);
		 return false;
		 }
	 });

 </script>
</head>

<body>
<a href="#popup-box" class="popup-window">Click</a>
  <div id="popup-box" class="popupInfo">
    <form>
      <label>ANYTHING</label></br>
      <input type="text" name="random"/></br>
      <button type="submit" name="button" class ="close">close</button>
    </form>
   </div> 

</body>
</html>

Edited by chauhanRohit, 27 June 2014 - 09:46 AM.

Hi, i have a query string id=5&start=0 that i want to remove from the url and return the url without it. I am searching a cleaner way for doing it. Here is an example :

Before : http://www.example.com/?a=3&id=5&start=0
After : http://www.example.com/?a=3

Before : http://www.example.com/?id=5&start=0
After : http://www.example.com/

Before : http://www.example.com/?alpha=1&id=5&start=0&beta=2
After : http://www.example.com/?alpha=1&beta=2

Thank you
Edited by Dareros, 11 August 2014 - 04:06 PM.

Hi,

When I run the following code...


<?php

if ($_GET['view'])
{
    echo 'test';
}


.. and then navigate to index.php?view=1 the above code outputs 'test', as intended.

The same goes for any non-zero value after the ?view=

However, navigating to index.php?view=0 does not work as intended because it assumes that the 0 is FALSE.

So if I modify the code, like this:


<?php

if ($_GET['view'] || $_GET['view'] == 0)
{
    echo 'test';
}


... it outputs 'test' even when view is not set, i.e. when I navigate to index.php without the ?view=.

Is there an easy way to force the value to behave as an integer instead of boolean? I have tried putting (int) before, as well as the intval() function, but neither seem to work.

Thanks.

I want the page to refresh on with the query string, but this isn't working for me...

Code: [Select]
header("Location: " . $_SERVER['php_self'] . "?" . $_SERVER['query_string'] );

Hey all,

When someone clicks a link, link looks like this:

http://site/homes/edit/1?image=true

Now on the same page, I have another link, which posts to the edit method as well, but no query string passed:

http://site/homes/edit/1

I want to check if image key is set to true and basd on that, determine which view to render for the user, that is, whether to display the edit form to edit post or the image form to edit image. But by using the get array to try to grab query string:

Code: [Select]
if($_GET["image"]=="true"){


I get the following error:

Message: Undefined index: image

It appears that $_GET is not successfully grabbing the parameter.

Thanks for response.

i want to change this code :

$result = mysql_query("select count(*) from 3gp where category='Bollywood'");

with this

$result = mysql_query("select count(*) from 3gp where category='$category'");

but i am getting error message "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/funxy/public_html/salmobile.info/3gp-videos.php on line 151". I am calling the page using http://abc.com/3gp-videos.php?category=Bollywood

Please help.


Code: [Select]
<?php
 

$db="funxy_db";

$conn = mysql_connect("localhost","funxy_saba","myasco001");
@mysql_select_db($db) or die( "Unable to select database, Please contact your administrator");









class Pager 
   { 
       function getPagerData($numHits, $limit, $page) 
       { 
           $numHits  = (int) $numHits; 
           $limit    = max((int) $limit, 1); 
           $page     = (int) $page; 
           $numPages = ceil($numHits / $limit); 

           $page = max($page, 1); 
           $page = min($page, $numPages); 

           $offset = ($page - 1) * $limit; 

           $ret = new stdClass; 

           $ret->offset   = $offset; 
           $ret->limit    = $limit; 
           $ret->numPages = $numPages; 
           $ret->page     = $page; 

           return $ret; 
       } 
   }  


 
    // get the pager input values 
    $page = $_GET['page']; 
    $limit = 3; 
    $result = mysql_query("select count(*) from 3gp where category='Bollywood'"); 
    $total = mysql_result($result, 0, 0); 

    // work out the pager values 
    $pager  = Pager::getPagerData($total, $limit, $page); 
    $offset = $pager->offset; 
    $limit  = $pager->limit; 
    $page   = $pager->page; 

    // use pager values to fetch data 
    $query = "select * from 3gp where category='Bollywood' order by id DESC limit $offset, $limit"; 
    $result = mysql_query($query); 

    // use $result here to output page content 


//my addition

//grab all the content

   //Custom Table Stsrt//
        $cols = 4; //number of coloms
        $i =1;
        echo "<table border=\"0\" cellpadding=\"2\" cellspacing=\"2\" width=\"100%\" id=\"table1\" bordercolor=\"#FFFFFF\" bgcolor=\"#FFFFFF\">"
             ."<tr>";

while($r=mysql_fetch_array($result))
{
   //the format is $variable = $r["nameofmysqlcolumn"];
   //modify these to match your mysql table columns
  
   $id=$r["id"];
   $name=$r["name"];
   $views=$r["views"];
   $url=$r["url"];
   $image=$r["image"];
   $category=$r["category"];
  
   
   
   //display the row
   
   
   
   $mybox = "
   
   <br>
   <a href='http://salmobile.info/3gp-videos-download.php?id=$id' class=\"classd\"><b><u>&nbsp;$name</u></b></a>
   <BR>
   <span class=\"text\"><img src=\"$image\" width=\"60\" height=\"60\"></span>
   <BR>
   <span class=\"text2\">Download This 3GP Video</span>
   <br>";


                if (is_int($i / $cols)){
                    echo "<td width='336' align='center' style=\"border-style: dotted; border-width: 0\">$mybox</td></tr><tr>";
                }else{
                    echo "<td width='336' align='center' style=\"border-style: dotted; border-width: 0\">$mybox</td>";
                }
             if ( $i / $cols == 3) echo "<td colspan='3' align=\"center\">









</td></tr><tr>";



if ( $i / $cols == 8) echo "<td colspan='3' align=\"left\">



 

</td></tr><tr>";


             $i++;
          //end if
       }//end while
       echo "</tr></table>";
       //Custom Table End//
   


//ends my addition



    // output paging system (could also do it before we output the page content) 
    if ($page == 1) // this is the first page - there is no previous page 
        echo "<font color=\"#FFB300\">Previous</font>"; 
    else            // not the first page, link to the previous page 
        echo "<a href=\"http://salmobile.info/index-" . ($page - 1) . ".html\" id=\"navigationURL\"><font color=\"#FFB300\">Previous</font></a>"; 

    for ($i = 1; $i <= $pager->numPages; $i++) { 
        echo " <font color=\"#FFFFFF\">|</font> "; 
        if ($i == $pager->page) 
            echo "<font color=\"#FFB300\"> $i</font>"; 
        else 
            echo "<a href=\"http://salmobile.info/index-$i.html\" id=\"navigationURL\"> <font color=\"#FFB300\">$i</font></a>"; 
    } 

    if ($page == $pager->numPages) // this is the last page - there is no next page 
        echo "Next"; 
    else            // not the last page, link to the next page 
        echo "&nbsp;&nbsp;&nbsp;<a href=\"http://salmobile.info/index-" . ($page + 1) . ".html\" id=\"navigationURL\"><font color=\"#FFB300\">Next</font></a>"; 
?>